Sunday, February 2, 2014

Two Quick Puzzles

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The following are two puzzles which look tough at first but have quick and really elegant solutions.

Q1: Ants on a wire:
A large number of ants are on a wire of length \(L\). All ants start moving randomly, either right or left with a fixed velocity \(V\). If they collide they turn around and move in the opposite direction. Ants at the ends of the wire fall off. What is the time taken for all ants to fall off the wire?

Q2: The Unruly Passenger:
Several passengers are in a queue to board a plane. The first passenger in the queue is an unruly one and chooses a seat at random. Subsequent passengers take their allotted seat if it is unoccupied or pick a seat at random if it is occupied. What is the probability that the last passenger gets to sit on his allotted seat?
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A1: This seemingly complex problem has an elegantly simple solution. The fact that they collide and turn around is the same as if they walked through each other! See figure below

Once this is insight sinks in, the average time taken for all ants to fall off the wire can be easily calculated. It is the same as the time an ant takes to move from one end of the wire to the other end. This works out to \(\frac{L}{V}\).

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A2: You absolutely do not want to consider the various ways a large number of passengers can fill up an equally large number of seats. Bear in mind that the only unruly passenger is the first one, and what we want to know is the probability that the last passenger gets to sit on his seat. The last passenger will face exactly two scenarios, either he gets his seat or not. He will get his seat if the first passenger picks his allotted seat which happens with a probability \(\frac{1}{2}\)

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  1. I am confused. The first passenger picks his seat with probability 1/n (n == num passengers), not 1/2, right?

    1. The first passenger picks his seat at random.

    2. Suppose there are 100 seats on the airplane. The unruly passenger sits in his assigned seat 1/100 times and chooses someone else's seat 99/100 times. So when we come to the last person, there's a 1/100 chance his seat is unoccupied, and a 99/100 chance that the unruly passenger caused a ripple effect and the last person is stuck in a seat not assigned to him. Right? Generalize for n.

    3. I'm not satisfied with A2 either. I offer the following argument: all of the passengers would behave in exactly the same way had the labels on the first and last passengers' seats been switched (or removed altogether). In other words, the given problem is symmetric with respect to these two seats. Hence the probabilities of the events "last passenger sits in first passenger's seat" and "last passenger sits in last passenger's seat" are equal. We know that one of these events must occur, since every seat other than those two must be occupied by the time the last passenger boards. Therefore the two probabilities are both 1/2.